Answer
Diverges.
Work Step by Step
Use the limit comparison test:
$\lim\limits_{x \to \infty} \dfrac{1/\sqrt {x}-1}{1/\sqrt {x}}=\lim\limits_{x \to \infty} \dfrac{\sqrt x}{\sqrt {x}-1}$
Now, apply L'Hospital's Rule:
$ \lim\limits_{x \to \infty} \dfrac{1/ 2 \sqrt x}{1/2\sqrt x}=\lim\limits_{x \to \infty} (1)=1$
Now, $\lim\limits_{a \to \infty}\int_{4}^{a} x^{-1/2} dx=\lim\limits_{a \to \infty}[2 \sqrt x]_{4}^a=\lim\limits_{a \to \infty}[2a^{1/2}-4]=\infty$
Thus, the integral diverges.